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(D) Let the position of the object be $u$ and the position of the image be $v$. The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

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$V$

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(D) Let the position of the object be $u$ and the position of the image be $v$. The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$,which implies $\frac{dv}{dt} = \left(\frac{v}{u}\right)^2 \frac{du}{dt}$.Here,the velocity of the object with respect to the lens is $v_{ol} = v_o - v_l = V - (-V) = 2V$. So,$\frac{du}{dt} = 2V$.At $u = -2f$,the image distance $v$ is found using $\frac{1}{v} - \frac{1}{-2f} = \frac{1}{f}$,which gives $v = 2f$.The magnification $m = \frac{v}{u} = \frac{2f}{-2f} = -1$.Thus,the velocity of the image with respect to the lens is $v_{il} = m^2 v_{ol} = (-1)^2 (2V) = 2V$.Since $v_{il} = v_i - v_l$,we have $2V = v_i - (-V)$,which gives $v_i = 2V - V = V$.

In the diagram shown,the lens is moving towards the object with a velocity $V \, m/s$ and the object is also moving towards the lens with the same speed. What is the speed of the image with respect to the earth when the object is at a distance $2f$ from the lens? ($f$ is the focal length.)

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